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Question

Two cars P and Q start with velocities u and u2, respectively at A but Q, 1 second later. They meet at B with velocities 22 m/s and 31 m/s, respectively. If their accelerations are a and 2a, respectively, find AB (in m)

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Solution

For car P,
v2=u2+2as
484=u2+2as (1)
For car Q,
961=u24+4as
3844=u2+16as (2)
From (1) and (2),
14 as = 3360 as = 240 (3)
Multiplying equ (1) by, 8 we get
3872=8u2+16as (4)
On comparing (2) and (4), we get
28=7u2 u2=4
u=±2
For car P,
v = u + at (By first equation of motion)
22 = 2 + at
at = 20
For car Q,
v = u + at
31 = 1 + 2a(t – 1)
31 = 1 + 2at – 2a
2a = 10
a=5 ms2
Put a = 5 ms2 in (3), we get
s=2405=48 m

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